Let $f(x, y) = \dfrac{x + y}{y - x}$ and $g(t) = (2 + t, 3 - t)$. $h(t) = f(g(t))$ $h'(t) = $
Answer: Formula The multivariable chain rule says that $\dfrac{dh}{dt} = \nabla f(g(t)) \cdot g'(t)$. The $g'(t)$ part is how much a change in $t$ will cause the input to $f$ to move, and the $\nabla f(g(t))$ part is how much $f$ will change in response to this update to its input. [What's the intuition behind the formula?] Applying the formula We want to find $h'(t) = \nabla f(g(t)) \cdot g'(t)$. $\begin{aligned} &g(t) = (2 + t, 3 - t) \\ \\ &g'(t) = (1, -1) \\ \\ &\nabla f = \left( \dfrac{2y}{(y - x)^2}, \dfrac{-2x}{(y - x)^2} \right) \\ \\ &\nabla f(g(t)) = \left( \dfrac{6 - 2t}{(1 - 2t)^2}, \dfrac{-4 - 2t}{(1 - 2t)^2} \right) \end{aligned}$ Substituting: $\begin{aligned} h'(t) &= \left( \dfrac{6 - 2t}{(1 - 2t)^2}, \dfrac{-4 - 2t}{(1 - 2t)^2} \right) \cdot (1, -1) \\ \\ &= \dfrac{6 - 2t}{(1 - 2t)^2} + \dfrac{4 + 2t}{(1 - 2t)^2} \\ \\ &= \dfrac{10}{(1 - 2t)^2} \end{aligned}$ Answer $h'(t) = \dfrac{10}{(1 - 2t)^2}$